A student walks (2.9±0.1)m, stops and then walks another (3.9 ±0.2)m in the same directionWith the given uncertainties, what is
1 answer:
Answer:
The smallest distance the student that the student could be possibly be from the starting point is 6.5 meters.
Explanation:
For 2 quantities A and B represented as
and
The sum is represented as
For the the values given to us the sum is calculated as
Now the since the uncertainity inthe sum is
The closest possible distance at which the student can be is obtained by taking the negative sign in the uncertainity
Thus closest distance equals meters
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Answer:
(a) E=0 : 0 cm from the center of the sphere
(b) E= 227.8*10³ N/C : 10.0 cm from the center of the sphere
(c)E= 911.25*10³ N/C : 40.0 cm from the center of the sphere
(d)E= 411.84 * 10³ N/C : 59.5 cm from the center of the sphere
Explanation:
If we have a uniform charge sphere we can use the following formulas to calculate the Electric field due to the charge of the sphere
: Formula (1) To calculate the electric field in the region outside the sphere r ≥ a
:Formula (2) To calculate the electric field in the inner region of the sphere. r ≤ a
Where:
K: coulomb constant
a: sphere radius
Q: Total sphere charge
r : Distance from the center of the sphere to the region where the electric field is calculated
Equivalences
1μC=10⁻⁶C
1cm= 10⁻²m
Data
k= 9*10⁹ N*m²/C²
Q=16.2 μC=16.2 *10⁻⁶C
a= 40 cm = 40*10⁻²m = 0.4m
Problem development
(a)Magnitude of the electric field at 0 cm :
We replace r=0 in the formula (2) , then, E=0
(b) Magnitude of the electric field at 10.0 cm from the center of the sphere
r<a , We apply the Formula (2):
E= 227.8*10³ N/C
(c) Magnitude of the electric field at 40.0 cm from the center of the sphere
r=a, We apply the Formula (1) :
E= 911.25*10³ N/C
(d) Magnitude of the electric field at 59.5 cm from the center of the sphere
r>a , We apply the Formula (1) :
E= 411.84 * 10³ N/C